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A Sisyphus Random Walk evolves as follows: With probability (r) you advance the position of the walker by +1. With probability (1-r) the walker resets at x0 = 0.

To simulate the probability of reset I use a Bernoulli Distribution:

t =  Prepend[RandomVariate[BernoulliDistribution[0.7], 9],0]

{0, 1, 1, 0, 1, 0, 1, 1, 1, 1}

According to this distribution the walk should evolve as follows:

srw = {0,1,2,0,1,0,1,2,3,4}

I am unsure what functions I should use to get the desired output.

We can iterate with FoldList:

data = {0, 1, 1, 0, 1, 0, 1, 1, 1, 1};



FoldList[#2 * (#1 + #2)&, data]

{0, 1, 2, 0, 1, 0, 1, 2, 3, 4}

Another approach is to accumulate consecutive runs of 1:

Join @@ Accumulate /@ Split[data]

{0, 1, 2, 0, 1, 0, 1, 2, 3, 4}

WARNING sorry, an error in this solution was just pointed out to me by another user. I am busy doing the necessary corrections.

A nice problem. Why not solve it analytically?

Let us use Latex to formulate the problem and write down the solution, and then move to Mathematica. Finally we discuss the results and generalizations.

Part 1: mathematical formulation of the problem

Let $w(t,k)$ be the probability that at time t(>=0) the walker is at position k.

Then let us derive the evolution equations. We do it carefully now because previously I had made an error here.

For $k>0$ we have the balance

$$w(t+1,k) = w(t,k) text{(old value)} - r w(t,k) text{(loss to k+1)}$$
$$+ r w(t,k-1) text{(gain from k-1)}-(1-r)w(t,k) text{(loss to 0)}$$

this gives the equation

$$w(t+1,k) = r; w(t,k-1), ;;;;;tge0, kgt0tag{1}$$

For $k=0$ we have

$$w(t+1,0) = w(t,0)text{(old value)}- r w(t,0)text{(loss to 1)} + (1-r) left(w(t,1)+w(t,2)+...right)text{(gain from all others)}$$

This gives

$$w(t+1,0) = (1-r) sum_{k=0}^infty w(t,k)$$

which greatly simplifies to the simple equation

$$w(t+1,0) = (1-r)tag{2}$$

Indeed, since at any time $tge0$ the walker must be at one of the locations $kge 0$ with certainty we must have
$sum_{k=0}^infty w(t,k)=1$.

In order to finalize the formulation of the problem we need initial conditions.

We assume that the walker at $t=0$ is in location $k=nge0$, i.e.

$$w(0,k)=delta_{k,n}tag{3}$$

where $delta_{k,n}$ is the Kronecker symbol defined as unity if $k=n$ and $0$ otherwise.

Part 2: solution with Mathematica

First we calculate w[t,0].

No calculation is necessary because (2) already gives the solution whih is constant for $tgt0$.

This result is independent of the specifiy starting point of the walker: if he would start at $k=0$ then, for t=1 he jumps to $k=1$ with probability $r$ which means that he remains at $k=0$ with probability $1-r$. If he starts at any other location he jumps back to $0$ in the next tmes step with probability $(1-r)$, hence we obtain the same result.

Now we turn to positions k>0.

(still to be checked)

From (1) we have

sol = RSolve[w[t + 1, k] == r w[t, k - 1], w[t, k], {t, k}]



(* Out[43]= {{w[t, k] -> r^(-1 + t) C[1][k - t]}} *)

Don't worry about the strange looking expression of the constant C[1] with an argument. This is exactly what we need to apply the initial condition (5).

Defining the auxiliary function

vc[t_, k_] = w[t, k] /. sol[[1]] t_, k]



(* Out[41]= r^(-1 + t) C[1][k - t] *)

condition (5) reads

sol1 = Solve[vc[0, k] == KroneckerDelta[k, n], C[1][k]]



(* Out[43]= {{C[1][k] -> r KroneckerDelta[k, n]}} *)

Hence the probability for the walker to be at location k at time t if he started at location n>0 (with the basic probability r of jumping one step to the the right) is given by

w[t_, k_, r_, n_] = r^t KroneckerDelta[k - t, n]



(* Out[50]= r^t KroneckerDelta[n, k - t] *)

Example 1:

n=1, r=1/2, w(t,k) is only different from 0 if k=t+1.

The first few values of these non vanishing Terms are

With[{n = 1, r = 1/2}, Table[w[t, t + 1, r, n], {t, 0, 10}]]



(* Out[58]= {1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, 1/512, 1/1024} *)

The complete probabilities form a two dimensional array, which might start like this

With[{n = 3, r = 1/2}, 

 tt = Table[w[t, k, r, n], {t, 0, 5}, {k, 1, 10}]]



(* Out[133]= {{0, 0, 1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1/2, 0, 0, 0, 0, 

  0, 0}, {0, 0, 0, 0, 1/4, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1/8, 0, 0, 

  0, 0}, {0, 0, 0, 0, 0, 0, 1/16, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 1/

  32, 0, 0}} *)

Visualization

ListPlot3D[tt, 

 PlotLabel -> 

  "Sysiphos random walk for r=1/2 starting at k=3nProbability w(t,k) 

for k>0 as a function of time", PlotRange -> {-0.1, 1}, 

 AxesLabel -> {"k", "t", "w(t,k)"}]

Part 3: discussion

Asymptotic behaviour for large times

The probabilities (6) and (7) at the origin approache a common value

$$w(ttoinfty,0) = frac{1-r}{2-r}tag{9}$$

As the walker must be initially either at the origin or not ("at home or out") the asymptotic value (9) holds for any initial state.

This means that the walker has asymptotically "gone out" with probability

$$w'(ttoinfty,0) = 1- w(ttoinfty,0) = frac{1}{2-r}$$

If the jump probability $r$ is small, it becomes equally probable that the walker is at home or out.

If $rto 1$ the walker is almost certainly out.

Next question: what is the asymptotic shape of the spatial profile? That is what is $w(ttoinfty,k)$ as a function of $k$?

My feeling tells me that is should be exponential ... like the pressure in the atmosphere.

Let's look at it qualitatively: if there were no jumping back, the walker would move away to $ktoinfty$ indefinitely. The jumping back leads to renewed filling from below of, in principle, any position.

What about $w(t,1)$? Let us try to calculate it exactly.